Two particles of masses m1, m2 move with initial velocities u1 and u2. On collision, one of the particles get excited to higher level, after absorbing energy ε. If final velocities of particles be v1 and v2 then we must have

Question

Two particles of masses m1, m2 move with initial velocities u1 and u2. On collision, one of the particles get excited to higher level, after absorbing energy ε. If final velocities of particles be v1 and v2 then we must have (A) (1/2)m1u12+(1/2)m2u22 -ε =(1/2)m1v12+(1/2)m2v22 (B) (1/2)m12u12+(1/2)m22u22 + ε =(1/2)m12v12+(1/2)m22v22 (C) m!2u1+m22u2-ε =m12 v1 +m22v2 (D) (1/2)m1u12+(1/2)m2u22 =(1/2)m1v12+(1/2)m2v22 -ε

Tardigrade Admin · Accepted Answer

Total initial energy of two particles =(1/2)m1u12+(1/2)m2u22 Total final energy of two particles =(1/2)m2v22+(1/2)m1v12+ ε Using energy conservation principle, (1/2)m1u12+(1/2)m2u22 =(1/2)m1v12+(1/2)m2v22 +ε ∴ (1/2)m1u12+(1/2)m2u22 -ε =(1/2)m1v12+(1/2)m2v22

Q. Two particles of masses $m_{1}$ , $m_{2}$ move with initial velocities $u_{1}$ and $u_{2}$ . On collision, one of the particles get excited to higher level, after absorbing energy $ε$ . If final velocities of particles be $v_{1}$ and $v_{2}$ then we must have

$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} - ε = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$

$\frac{1}{2} m_{1}^{2} u_{1}^{2} + \frac{1}{2} m_{2}^{2} u_{2}^{2} + ε = \frac{1}{2} m_{1}^{2} v_{1}^{2} + \frac{1}{2} m_{2}^{2} v_{2}^{2}$

$m_{!}^{2} u_{1} + m_{2}^{2} u_{2} - ε = m_{1}^{2} v_{1} + m_{2}^{2} v_{2}$

$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2} - ε$

Q. Two particles of masses m1​, m2​ move with initial velocities u1​ and u2​. On collision, one of the particles get excited to higher level, after absorbing energy ε. If final velocities of particles be v1​ and v2​ then we must have

21​m1​u12​+21​m2​u22​−ε=21​m1​v12​+21​m2​v22​

21​m12​u12​+21​m22​u22​+ε=21​m12​v12​+21​m22​v22​

m!2​u1​+m22​u2​−ε=m12​v1​+m22​v2​

21​m1​u12​+21​m2​u22​=21​m1​v12​+21​m2​v22​−ε

Q. Two particles of masses $m_{1}$ , $m_{2}$ move with initial velocities $u_{1}$ and $u_{2}$ . On collision, one of the particles get excited to higher level, after absorbing energy $ε$ . If final velocities of particles be $v_{1}$ and $v_{2}$ then we must have

$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} - ε = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$

$\frac{1}{2} m_{1}^{2} u_{1}^{2} + \frac{1}{2} m_{2}^{2} u_{2}^{2} + ε = \frac{1}{2} m_{1}^{2} v_{1}^{2} + \frac{1}{2} m_{2}^{2} v_{2}^{2}$

$m_{!}^{2} u_{1} + m_{2}^{2} u_{2} - ε = m_{1}^{2} v_{1} + m_{2}^{2} v_{2}$

$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2} - ε$